The question requires us to calculate the amount of solution that could be prepared from 1.85 g of silver nitrate (AgNO3), considering a 5.0% solution.
Considering that 5.0% solution corresponds to a solution where there are 5g of AgNO3 in 100 g of solution (% m/m), we can write:
5.0 g of AgNO3 ---------- 100 g of solution
1.85 g of AgNO3 --------- x
Solving for x, we have that 37g of a 5.0% solution could be prepared from 1.85g of AgNO3.
